Finding the value of x at which the tangent to the curve is parallel to the x axis I have thoroughly searched up how to attempt this question. However, I am not sure if my answer is correct or if I even attempted the question correctly. Assistance would be greatly appreciated!

Calculate the value of x at which the tangent to the curve is parallel
  to the $x$ axis for the curve $y=3x^2+5x-2$

The value I got for $x$ is $0$.
How I did it:
the expression for the gradient $\frac{dy}{dx}= 6x+5$
then, $6x+5=0 \Leftrightarrow 6x=-5 \Leftrightarrow x=-0.83$
or if that was wrong I also tried
$$6x+5x
\Leftrightarrow6x+5x=0
\Leftrightarrow11x=0
\Leftrightarrow x=0$$
I dont think I did it properly. Any suggestions?
 A: The tangent to the curve is parallel to the $x$-axis when the tangent has slope equal to zero i.e. when the derivative equals zero. You properly computed the derivative and solved for the value of $x$ when $\frac{dy}{dx}=0$ in your first approach. The value of $x$ is $-5/6$.
A: Your first attempt is correct as the derivative of $y=3x^2 + 5x - 6$ is $y'=6x+5$
When the tangent to the curve is parallel to the $x$ axis then $y'=0$
Hence $$y'=6x+5 \Leftrightarrow 6x+5=0 \Leftrightarrow x = \frac{-5}{6}$$
A: All correct in your first part. Stated condition occurs when derivative vanishes.
A: You found the correct point $x = -5/6$. There the tangent has slope $0$ and is thus parallel to the $x$-axis.

(Large version)

$$6x+5x \Leftrightarrow6x+5x=0 \Leftrightarrow11x=0 \Leftrightarrow
> x=0$$

The first equivalence is not true. You can not infer $6x + 5x = 0$ from the expression $6x + 5x$.
It is also unclear how you arrive at $6x + 5x$ in the first place.
A: Since $y(0)=y(-5/3)=-2$, the first coordinate of the vertex is the average of $0$ and $-5/3$, namely $-5/6$.
